Karma-Audio.dk

 

Teknik Radio RÝr

 

Effects of AC Heating Power Applied to Directly Heated Triodes

(c) Copyright 1999, Steve B. All Rights reserved. Revision 2.

This report examines the effects of using AC to heat the emitting structure (filamentary cathode) of a Directly Heated Triode. I will concentrate on these effects as applied to a single ended amplifier stage, as some of these effects are automatically cancelled in a push-pull arrangement. I will use an 801 thoriated tungsten filament device as a working example. The principles discussed in this report are applicable to all filamentary cathode structures.

Three major effects occur:

  1. Line frequency hum, primarily induced by imbalance in the filament circuit wiring, the transformer used to power the filament, coupling into the signal leads, etc. This is typically reduced or eliminated using a "hum balancing pot". This is a potentiometer connected across the filament whose wiper arm connects to the DC return of the system. This is fairly commonly utilized, and affects the line frequency component (and odd harmonics). No more about this will be mentioned in this report, as the principles are well established. However, the schematics included herein include this technique.
  2. Hum produced from within the filamentary structure. This is primarily caused by the lack of "infinite mass" of the filament. Since heating power is symmetric for each half cycle of applied voltage, the principle component of this hum is twice the line frequency (and even harmonics of the line frequency). Physically, the filament minutely heats up and cools down at twice the line rate, due to AC applied to the filament. The next section of this report covers this effect and its reduction.
  3. Signal modulation of the hum component. The effect is an intermodulation of the applied signal at twice the line frequency. This effect applies equally to single ended and push-pull directly heated devices. The effect is not a hum (low level) effect; rather, it is a high level distortion effect. For example, with a 60Hz line frequency, if you were to introduce a pure 1000 Hz signal into the amplifier, you will get not only an amplified 1000 Hz output (and some level of harmonics at 2000 Hz, 3000 Hz etc) but also a hum modulation set of components. The primary components in this example are at 880Hz and 1120Hz. As this is an intermodulation effect, there will also be components at many other frequencies, for example 760 Hz (1000-240 Hz), 1640Hz (2*1000-3*120) etc. Note that IM products, unlike harmonic components, are not sonically related to the signal. The third section of this report discusses the magnitude of this effect, and the conditions needed to minimize it.

The principles discussed in this report have been experimentally verified on the following devices:

801: Low MU Thoriated Tungsten filamentary heating structure.

841: High MU Thoriated Tungsten filamentary heating structure.

6B4G: Low MU Oxide Coated filamentary heating structure.

1U4 (triode connected): Medium MU small mass filamentary heating structure.

2. DHT AC Powered Hum Control

This section covers low level hum reduction associated with line powered DHT filaments. I will assume the line frequency component has been balanced out with the standard "hum pot". However, even when you have done this, the left over hum, primarily at twice the line frequency, may still be objectionable. Since it is at twice the line frequency, it is often assumed to be associated with the bias or HT circuitry. Attempts to better filter these sources sometimes makes things worst. In this section, we will see why that is so.

As mentioned above, the slight heating and cooling every half cycle of the AC input frequency is the source of this hum component. As it turns out, the phase of the hum components is very fortuitous. Namely, a full wave rectified signal producing a negative voltage (such as would be used for bias) injected into the grid reduces the hum, and a full wave rectified producing a positive voltage (such as used for the HT) injected into the plate also reduces the hum. This suggests that a properly phased and "inadequately" filtered source is actually beneficial in reducing this hum component. More on this later.

We will cover the magnitude of the effect in this paragraph. To investigate this, I used a filter bank to obtain a set of signals locked to the incoming line frequency. I then adjusted the phase and amplitude of each signal to see what hum reduction could be obtained. My "mains" frequency is 60 Hz. Those with other mains frequencies should translate the frequencies accordingly. I set up an experimental 801 amplifier. This was an RC coupled circuit. I also slightly "starved" the filament by running it on 6.3V instead of 7.5V. (Actual filament voltage was 6.62VAC). The HT was 600VDC VERY well filtered (residual AC component was 10 microvolts). Bias source was obtained from a 5651 filtered for a residual AC component of about 3 microvolts. This was done so that I could independently study the effect of the DHT hum components. The circuit was biased at about -26.5V, and the plate load was 27k, producing a circuit gain of about 6. The filament was balanced with a hum pot. The residual hum was 342 millivolts at the plate. (Note that if this were an output stage, thru the transformer impedance transformation and load, this would translate to a few mV, which is typical for SE DHT stages). Injecting a properly phased 120Hz component allowed a hum reduction of 18 dB. Adding in a properly phased 240Hz component increased this hum rejection to 25 dB. Adding in a properly phased component at  360Hz increased this rejection to 29dB. At this point, that major hum component left was 180Hz, as there is a phase difference due to the mass (capacitance) of the filament structure) that causes the 60 and 180 components to be slightly out of phase with each other. Adding in properly phased 180Hz increased the rejection to about 31 dB. This rejection remained stable over time and with cycling of power on/off. As long as the inserted cancellation components tracked the magnitude of the "line", the rejection stayed within a dB or so from a line input of 110VAC to 130VAC. Using the filter bank approach, the best I was able to achieve was 36 dB (120, 180, 240, 360, 480, 540, 600 Hz filters).

A practical circuit to do this would consist of perhaps 3 LC bandpass filters: tuned to 120Hz, 240Hz and 360Hz. Then, by slightly adjusting the resonant frequency, the phase can be altered, and by adjusting a "summing node" resistor the amplitude could be controlled.

There is a MUCH simpler "compromise" network that can provide very effective control. As I mentioned above, the phase of the induced hum components is such that a negative full wave rectifier injected on the grid or a positive full wave rectifier injected on the plate is of the correct phase to cancel this hum. Therefore by adding THREE COMPONENTS to the typical circuit, the DHT hum introduction can be cancelled. Here's the circuit:

The "left half" of the schematic is rather conventional. It does include the standard "hum pot". The "right half" circuit shows the introduction of two small diodes and a second pot to correct for the DHT induced hum. Using this circuit, I was able to reduce the output side hum from 342 mV to 29 mV, for almost 22 dB reduction in overall hum level. Notice that this compromise circuit provides most of the advantage of a very complex circuit with almost no added complexity. For the crowd who shun the use of the small silicon signal diode, a 6AL5 or 6H6 could be used instead. By the way, I used a "high quality" 10 turn pot for the second hum reduction potentiometer. There did not seem to be much difference with the type of diode, although depending on the tube and "layout" sometimes 0.02uF across each of the diodes helped.

This technique is also applicable to "cathode" bias techniques. Here's a sample circuit, again using an 801 as the example tube:

Notice that this is no more complicated than the fixed bias example. The cathode bypass should be "effective" well below the "mains" frequency", which is usually the case anyway. If the filament transformer has a center tap, don't use it.

This technique can be extended to include the entire power amp "front end". Here is an example of my entire driver amp with hum compensation:

In this example I added one more LC, roughly tuned to 240 Hz in a high pass configuration. The compensation was "injected" into the 6SN7 stage. Notice that the "phasing" is the same. Since this is additional injection is band pass (filtering 240 Hz, 360 Hz etc), I could bypass the first canceller with a 220 nF capacitor, providing some additional higher frequency line noise rejection. The overall output ripple from this circuit is about 12(!) millivolts, for a overall hum reduction of 340/12 or 29 dB reduction.

BTW, this driver provides about 40 dB of gain, and makes a very nice sounding power amplifier driver.

Other DHT Types

The 841 has essentially the same filamentary structure as the 801, and compensated almost exactly the same, when you account for the difference in gain (and difference in load, bias etc). The 6B4 has a slightly less massive filament structure, and the phase and amplitude components needed for cancellation were different, but about the same reduction was able to be effected. The 1U4, with its tiny filament didn't compensate quite as well, allowing only a 15 dB improvement.  I did not try the filter bank approach with the 1U4; this may allow more improvement. In case it wasn't obvious, DON'T EVEN THINK about applying 600 volts to a 1U4. The phase differences can be handled by either changing the "zero" (changing the 2.2uF in the circuit above), or adding a "pole" by adding a capacitor to ground at the wiper of the pot etc.

Other Methods

Also as mentioned above, an inadequately filtered HT and/or bias supply is actually beneficial. However, the phasing is such that the required filter is a power supply consisting of ONLY an RC filter. LC or CLC filters produce excessive phase shift and won't properly compensate. Using an RC filter, simply adjust the output C for minimum hum! There will be a definite "minimum"; increasing the C above this point will cause the output hum to increase, as the supply "ripple" is too low to compensate. Notice that a full wave section, not half wave nor doubler should be used.

3. HUM MODULATION Effects of the DHT

As mentioned in the introductory section, there is also an intermodulation effect caused by the minute temperature changes in the filament. This incrementally changes the transconductance of the device, leading to a modulation effect. This effect will be observable in SE and PP applications. In this section, we will investigate the magnitude of this effect and determine how it might be possible to minimize the effect.

To study this effect, I used the same circuit as shown in the schematic above. I applied various levels of 1000 Hz signal to the input of the amplifier, and measured the resulting component at 880 Hz as an indicator of this effect. The bias voltage was carefully adjusted to 26.5V bias, and the Filament voltage was 6.6V

For this test, I will "table" the INPUT peak signal level vs % modulation.

Input Peak Level vs % Modulation
Input Level (Pk) Hum IMD %
10 0.12%
15 0.13
20 0.14
21 0.142
22 0.144
23 0.16
24 0.20
25 0.24
26 0.28
27 0.31
28 0.34
29 0.35
30 0.35

The peak filament AC is 4.6V peak (half of 6.6*1.4). If you subtract this from the 26.5V bias you get 21.9V. Notice that in the table, the distortion more or less "linearly" increases with signal level (which, incidentally, is NOT normally characteristic of IM) until you get past 22V peak signal, then the hum modulation increases rather rapidly, reaching a maximum at about clip point (which is evident at 28-30V peak input).

This suggests that the way to keep hum "modulation" effects under control is to establish the bias conditions such that the bias is greater than the signal level + the peak of the AC filament voltage.

The hum cancellation discussed in section 2 did not affect the hum modulation by more than 1 dB, when cancelled in either the grid or the plate circuit of the 801. Not shown on the schematic above, but this stage is driven by a 6SN7 "input stage". When the hum cancelling signal is injected into the a small 10 ohm "sampling" resistor in the cathode of the 6SN7, there was no effect of the hum cancellation on the hum modulation IMD. This suggests that the small effect observed in the 801 grid is due to the slight variation in plate resistance of the 6SN7, and the slight effect when injected at the 801 plate is due to the slight variation in plate resistance of the 801.

Steve

 

 

 

Directly Heated Triodes operated with lower voltage on the filaments - Updated.


One of the interesting effects I ran across was the effect of operating directly heated triodes at reduced filament voltages. As it turns out, this drastically reduces the distortion produced in the device. This effect seems occur on all forms of directly heated tubes, from power tubes to small signal tubes, to "battery" tubes. The effect, for reasons shown later, can be exploited in small signal applications, but is more difficult to realize in power applications.

Here are some results of data collected in the lab:

O1A. Plate load=100k, Plate supply=330V, Quiescent plate voltage=132V.

Filament   Vout(rms)   THD
5v       10v       0.19%
20v 0.38%
30v 0.58%
40v 0.79%
50v 0.90%
3.15v 10v 0.03%
20v 0.06%
30v 0.10%
40v 0.20%
50v 1.5%

Even large tubes show this effect. Here's data on a 1619 (similar to a 6L6, except it has a directly heated 2.5v filament instead of the heater-cathode.of the 6L6. The data is for a 330v supply, 25k load resistor, with the bias adjusted for 150 volts on the plate.

Filament  Vout(rms)  THD
2.5v   10v       0.70%
20v 1.60%
30v 2.50%
40v 3.60%
50v 5.00%
1.4v 10v 0.23%
20v 0.40%
30v 0.55%
40v 0.80%
50v 1.10%

Here is data for a type 26. This was the "best" of  seven tested 26 devices. The conditions are 330 volt plate supply, 100k plate load, quiescent voltage adjusted for 132 volts.

Filament  Vout(rms)  THD
1.5v     10v       0.28%
20v 0.58%
30v 0.90%
40v 1.20%
50v 1.50%
0.8v 10v 0.03%
20v 0.06%
30v 0.10%
40v 0.18%
50v 0.34%

This effect occurs whether the filament is AC powered or DC powered, and is not hugely sensitive to the voltage. Here's data for a 1U4 battery tube illustrating the effect of filament voltage. The conditions are 220 volt plate supply, bias adjusted for about 109 volts quiescent. The plate load was 200k, and the output level was maintained at 20 vrms.

Filament  THD
1.4v     1.90%
1.3v 1.85%
1.2v 1.80%
1.0v 1.50%
0.9v 1.10%
0.85v 0.60%


Effect as a Function of Voltage - Harmonic Components

One of the possible problems with the application of this effect is a potential cancellation of the second harmonic component, at the expense of the other harmonic components. This does not have to be the case. In the picture below, a 6B4 was characterized as a "small signal tube" - exotic, but possible. The picture shows that the higher order components are reduced even more than the second harmonic, until you get to very low filament voltages. Here's the picture:


The effect explained

In order to understand what was going on, I built up a curve tracer to view the tube characteristics. I used the 01A as the test device, as the effect with this device is representative of the class. The curves shown were obtained by tracing the curves from the scope, then entering the values into an Excel spreadsheet. (There was a previous version I put on the web page that was a "picture" of the tracings. It was ugly enough to redo in the format now shown). The curves below both use 2 grid volts per step. The 0, -10, and -20 volt biases are shown in red for better clarity.



This is VERY interesting. The "starved" condition now includes a wide range where the mu, transconductance, and plate resistance are essentially constant for a wide range of loads... The "normal" characteristics have the familiar "plate resistance varies and gm varies, but mu is constant" shape. The starved characteristics clearly show a flattening out of the plate current (due to the starvation), but the range of constant gm and constant mu is also evident. The flattening out of the characteristics also implies that this effect would be harder to implement in a power amplifier situation.

The "data sheet" values for this tube are rp=10k, gm=0.8mS, mu=8. Reading the values from the above tube curves, near the datasheet operating point, I get rp=8.8k, gm=0.95mS, mu=7.5. Reading the values for the starved filament condition, again near the same operating point, the values are rp=13.5k, gm=0.63mS, mu=8.5. In a real circuit, there is a measureable increase in circuit gain as the filament voltage is lowered, consistent with the tube curves. The big difference, however is still the wide range over which the tube parameters are essentially constant.

So maybe we have explained the mystery; by starving the filament in DHT devices, the tube characteristics show a substantial region of not only constant mu, but constant gm and constant rp, thereby essentially removing all distortion causing mechanisms.

Incidentally, I have been "life testing" a 1U5 running under starved filament conditions, to make sure that it is possible to exploit this effect without harming the lifetime of the tube, due to "cathode" stripping. Although this would normally consume much time, preliminary data (after slightly over 2 weeks time so far), there has been no degradation in the transconductance measured of either the starved tube, or on the"normally operated" (control) 1U5 concurrently being tested.


For further study: Indirectly heated triodes do not have this effect with one exception: the 27.

Steve

 

 

Purpose

The purpose of this article is to describe different ways in which you can hook an output transformer and plate (anode) circuit together, along with the relative advantages and disadvantages of each. Five different structures are described: standard transformer coupling, parafeed, resistively loaded stage capacitively coupled to the output transformer, tube (valve) based constant current source load capacitively coupled to the output transformer, and solid state (MOS FET) constant current source load capacitively coupled to the output transformer.

These configurations apply to both single ended and to push pull output stages. I will use single ended examples, followed by an illustration of how this is also applied to push-pull amplifier stages.

The 5 Output Structures:

Transformer Coupled Stage

The first illustrated example is standard transformer coupling. In the single ended situation, the major disadvantage of this method is the average DC current that flows in the circuit must be supported by the transformer, making the transformer relatively large so that it does not saturate due to the DC only. This is probably the most common output coupling method, as it is relatively simple. At quiescent conditions, the power supply voltage appears on the plate (minus a little loss due to the IR loss in the transformer. During signal conditions, the output swings above and below the supply voltage.

All the other output structures below will be compared to this output coupling mechanism.

Parafeed

The "parafeed" (parallel feed) uses a large choke to provide the DC voltage to the anode. This choke must be large in value, and provides a relatively high impedance at audio frequencies. The output transformer is capacitively coupled to the anode. Sometimes the capacitor is inserted between the anode and the transformer (as shown in the schematic above), and sometimes the transformer is connected directly to the anode, and the capacitor inserted in the ground lead.

The advantage of this kind of mechanism is the choke and the transformer can be individually optimized: the choke for saturation capability, and capacitance; whereas there is no DC in the transformer, so it can be made relatively smaller, providing both lower capacitance and leakage inductance with respect to its primary inductance. This is a long winded way of saying that it is possible to achieve wider frequency response.

There are four other advantages to this circuit. First, the capacitor inserted allows a low frequency "extension" due to the resonance between the transformers inductance and the capacitance value. The second is tha additional low frequency poles can be more easily controlled (as the capacitor and the inductance of the choke can be used as additional "degrees of freedom" in the design). The third is reduced "hum". In a normal transformer circuit, the hum voltage forms a voltage divider between the plate resistance and the load. For low plate resistances, the power supply ripple is coupled to the output. In the parafeed circuit, the power supply ripple forms a voltage divider across the choke, and is not coupled to the output. Thus, the power supply characteristics are more isolated from the output. The fourth advantage of this circuit is since the transformer has no high voltage DC on it, the transformer can be replaced with an "autoformer" (a single tapped winding) allowing the output autoformer to be further optimized.

Like the transformer coupled circuit, the choke feed in the parafeed circuit allows the output to swing above and below the power supply voltage.

The disadvantage of the parafeed is the size, weight and cost associated with TWO rather large magnetic elements in the system. The second disadvantage of the parafeed is that there is a capacitor in the signal path, and this capacitor is handling a relatively large AC current, potentially adding additional non-linearities to the circuit. It doesn't matter where the capacitor is positioned in that series circuit: the signal current still flows through it.

Resistor Load, Capacitively Coupled to the Output Transformer

In this topology, the output tube (valve) is powered by a large resistor fed from the power supply, and the output is capacitively coupled to the transformer. This is very much like a normal "preamp" stage, except the values are somewhat different.

This type of configuration requires a much higher power supply voltage, since the power supplied to the output stage is supplied through this load resistor.

At this point, it is probably worthwhile to consider an example. Consider a SE output stage consisting of a type 10 (or type 801 if you like). This is a DHT device. I'll consider an operating point of 350V and 11 mA (approximately -30V grid bias). With a normal transformer output configuration, the HT voltage would be about 360 volts (the extra 10 volts to account for the transformer winding resistance). I'll also consider a 12k AC load. This will provide a blazing 700 mW of power.

For a resistor load instead, I'll consider using a 35k resistor to a supply voltage of 735 volts. At quiescent, 11 mA is dropped across the 35k resistor (385 volts), leaving 350 volts to be applied to the anode. With the transformer capacitively coupled, the load on the tube is 35k in parallel with 12k or about 9k. Instead of getting 700 mW, we will get about 500mW. Also the 35k resistor will be dissipating over 4 watts!   

Advantages of this configuration: Like parafeed, there is no DC in the transformer, so a smaller unit can be used. Also the big bulky choke is replaced by a relatively "small" power resistor. From a "load" viewpoint, the 35k looks to be in parallel with the anode resistance, so the damping is slightly better. There is no issue of inductor saturation either. This configuration offers the widest frequency response (LF is improved as the transformer is looking at a lower resistance than the anode resistance alone, and there is no parallel inductance to limit the LF. HF is improved as there is no real capacitance due to the choke.). Also, like parafeed, hum is reduced by the effective voltage division between the "pull up" resistor and the anode resistance. The comments relative to an output "autoformer" also apply.

Disadvantages of this configuration: More "heat" is generated, power output is somewhat lower, and the power supply must be much higher! The output cannot swing "above the supply" so the supply must be great enough to accomodate the expected swing. This is probably a long way around of saying you need to plot 2 separate load lines; the AC load line and the DC load line. (You, of course KNOW how to do that from reading the "Of Loadlines..." series on these pages).

You will almost *never* see this configuration used, because of the mentioned disadvantages. [I would like to add that I find something magical about the sound of the resulting amplifier, in spite of its disadvantages].

Constant Current Loading

The idea behind the constant current load is that with a constant current load, the valve is almost perfectly linear (near ZERO distortion). However that discounts that the AC load is still the transformer reflected impedance. However, in principle, it provides essentially a parafeed connection topology with a more "perfect" choke... it has a high AC impedance and still powers up the circuit.  The more "perfect" nature is that the high impedance does not change with frequency as it does with a choke. I will describe 2 implementations; one using a tube constant current source, and one using a solid state implementation.

A Tube Constant Current Load

The circuit shown above provides about 11 mA with 385 volts across the constant current element (6SN7). The effective impedance is about 50k or so. Since the 6SN7GTB can handle 450 volts and 11 mA within its ratings, it would operate OK in our example above. (In fact, for 350 anode quiescent voltage, the HT supply could be 800 volts).

Advantages: Like the previous 2 topologies, there is no DC in the transformer. The frequency response will be good as well, since there is no loss in effective impedance at low frequencies. The power will be slightly less than with parafeed (as the resistance at midband will be lower than with the choke), but greater than the resistor pull up configuration. Less size and weight than the parafeed. Also, unlike the resistor example, as the output tube swings towards cutoff, (higher than quiescent voltages) the current tends to remain constant, so the current can be delivered to the load. With the resistor load above, this is not the case. This means that in principle, to deliver the same power as with the resistor load example, a lower HT supply would work successfully. (The example shown works to about 80 volts drop across the 6SN7. For the "blue example" above, rather than 735 volts, "only" 600 volts would be needed for the HT supply). Note that the comments relative to using an output "autoformer" also apply to this circuit as well.

Disadvantages: Like the resistor load example, much higher voltage is required. An extra tube is needed. Lower efficiency than the parafeed or transformer example.

There is a nasty disadvantage to any constant current source load. This type of topology is really only applicable to triodes. Since a pentode presents a fairly constant current sink, trying to power it from a constant current source will be fairly unsuccessful, sort of like the unstoppable object hitting the immovable object. Also, cathode bias circuits (with bypass capacitors) are not very successful, since for DC bias purposes, the cathode resistor presents a high DC resistance equivalent at the anode, and the quiescent point is not very stable.

A Solid State Constant Current Load

The circuit shown above will provide about 11 mA for about 385V drop. With the parts shown, the effective impedance is about 400k. (So it is a much more "constant" constant current source). It is effective for voltage drops of 20 volts to about 1000(!) volts across the constant current circuit, so it allows for 500 volts "quiescent" conditions across the current source. (In the example above, the HT supply could approach 1kV). The transistors would need to have heat sinks.

Advantages: Essentially the same as above, except with the higher impedance of the circuit, the power available approaches that of the transformer coupled circuit alone. In the "blue example" above, the HT supply would need to be only 550 volts. Notice that there is a substantial range between the minimum required supply voltage (550 volts) and the maximum possible  voltage before blowing the transistors (nearly 1kV).

Disadvantages: Similar to above. There might be considered to be 2 additional disadvantages: a heat sink is required (and the cases of the transistors are at high voltage), and the "solution" is solid state, which might not be satisfactory to some potential users.

Push Pull Equivalent Structure

Push Pull Example Discussion

Push pull offers an advantage that SE amplifiers don't offer relative to the output transformer. Since the DC current on each side is the same, the effective DC in the transformer core is cancelled, so there is no need to provide for a large DC component. That usually means push pull amplifiers are almost always "transformer only" coupled.

However, some mismatch (and thus unbalanced DC) can occur in the circuit, leading to (perhaps) a change in characteristics of the amplifier as a function of loudness or over time. This can be avoided by using any of the same techniques discussed in the SE section above to provide the DC paths to the output tubes (valves). In the schematic above, I used the transistor constant current source example. Notice that the same "disadvantage" as the above applies... the HT voltage must be much higher to allow the tube to swing "above" its quiescent point, just as noted above.

There is also an equivalent additional advantage to this topology. In addition to being immune to tube unbalance, an autoformer approach can be used. In this case, a single winding is again used, but it is center tapped (and this is the ground connection), and tapped at 4 ot 8 or 16 ohms away from the center tap. This looks "odd" but works just as effectively. In ASCII art, this winding is.....

   Plate     G Out Plate

     |          |  |       |

     UUUUUUUU

     ---------------

-Steve

 

 

Tube Based Voltage Regulators

Purpose

The purpose of this report is to provide a primer on vacuum tube based voltage regulator circuits. We will do this in several parts:

  • Part 1 discusses several one tube based voltage regulators.
  • Part 2 Adds a voltage reference device to our simple voltage regulator to provide line as well as load regulation.
  • Part 3 Introduces an "Error Amplifier" Stage.
  • Part 4 describes circuits with multiple control tubes.
  • Part 5 discusses shunt regulators.

For all these circuits, we will use the same rectifier and filter section, so that you can see the difference in performance of the various topologies covered in this primer.

(c) Copyright 2000, Steve  All rights reserved.


Part 1 - One Tube Regulators.

One tube regulators provide a form of electronic filtering, and some "load" regulation; that is, the output voltage remains fairly constant as the load on the regulator changes.

Since these simple regulators do not have a "reference", they can not provide "line" regulation. That is, as the AC "mains" voltage changes, the output voltage will change as well. The lack of line regulation has one odd advantage: at low line voltages, the output  drops along with the line, so the voltage drop across the tube doesn't drastically change. In part 2, when we introduce a reference, this effect will become more apparent, since the output wants to remain constant even at low line voltages. The consequence of that additional regulation is the possibility of the voltage across the tube becoming too low to regulate. These one tube regulator circuits do not have that problem. (see basics, below).

In all the examples shown, I will describe the "ripple" emerging from the supply, and graph the output voltage as a function of load. I used 60Hz "mains" in this example. Those with 50Hz mains will observe slightly more ripple for the values shown.


The Rectifier and Filter circuit used in the examples

Here is the basic power supply and filter circuit.

This circuit provides about 294 volts at a 50mA load and nominal line. It also produces about 800mV of ripple. At 20% low line, this voltage drops to 232 volts, at -10% low line, the voltage is 260 volts. At +10% high line, the output goes to 320 volts, and at 20% high line, the output climbs to 355 volts. The voltage vs load current looks like this:

Notice that at load currents below 40mA, the voltage rises rapidly. This is due to the LC filter. At those currents, the critical current for choke input filtering is not satisfied. This is done purposely so we can see the effects of different regulators.


Regulator Basics

The purpose of a voltage regulator is to maintain a relatively constant voltage no matter how the load or power line changes. For the initial set of regulators discussed in this part, the regulator can provide some "load" regulation. This means that the voltage will remain relatively constant as the current drawn from the supply changes. A figure of merit for the "goodness" of a regulator is its output resistance. Ideally it should be zero. For the regulators in this part, it will approach 1/gm of the tube being used.

A good regulator should maintain the same output voltage when the line or "mains" voltage changes. This usually requires some form of "reference" tube in order to do this. For the regulators in this part, there is no "line regulation" capability. This means the output voltage will change when the line voltage changes. In part 2 of this series, we will discuss how to provide line regulation.

A good regulator should also improve the "output ripple" (hum) emerging from the rectifier and filter. A figure of merit for regulators is the "ripple reduction" provided. For the regulators discussed in this section, ripple reductions will be between 3 and 20 to1 ripple reduction. Part 3 will introduce improvements that allow better ripple reduction.

A typical characteristic of regulators is the ability to set an output voltage, independent of the rectifier and filter voltage. (Well, set to any voltage somewhat BELOW that produced by the rectifier and filter anyway). This is usually done by a potentiometer or voltage dividing set of resistors to the grid of the regulator tube. The output from the regulator is then typically the cathode of the tube. Since the grid doesn't draw current, this voltage divider does not need to be high power.

The tube intended to provide the regulation is called the "pass" tube. Good characteristics for pass tubes include high transconductance (providing low output resistance), high current capability (to allow maximum current capability from the regulator), and high power dissipation. The dissipation is the voltage dropped across the tube multiplied by the current delivered. For example, if the rectifier and filter produce 400 volts, and we desire to produce 250 volts at 100 mA, there is 150V*100mA=15 watts dissipated in the pass tube. Notice that power dissipated in the pass tube INCREASES as the output voltage is set LOWER, for any output current. We will discuss this further in part 2.


The Tube used in this study.

For this primer, I chose to use the 6KG6/EL509 as the regulating device. This is a beam power tube. In some of the tests, it will be triode connected, providing better performance than 2 sections of 6AS7/6080. It also allows us to use it in a "pentode" connection, to see if that is a help or hinderance in power supply operation. (Common wisdom suggests low mu, low plate resistance triodes).

The tube is capable of 35 watts plate dissipation, and 7 watts screen dissipation (for a triode mode connection capability of about 40 watts), 500mA maximum current, 18mS transconductance, and 900 volt plate rating.


Self Biased Triode Mode Regulator

The simplest regulator is a self biased series pass regulator.

I've shown two conditions: one delivering 200volts, the othher delivering 250 volts. The regulation characteristics of this circuit are:

Notice that the voltage climbs pretty fast at low currents, and the load regulation is not quite as good as the choke input filter. What's the advantage? This circuit reduces the output ripple from 800 mV to about 150mV (250 volt case) or 200mV (200 volt case). When set for 250 volts, with -20% to +20% variation in line, the output voltage changes from 192 to 306 volts. When set for 200 volts, with -20% to +20% variation in line, the output changes from 154 to 245 volts. (at 50mA load).

Also, ripple can be further decreased by increasing the capacitor value on the grid of the tube.


Fixed Biased Triode Mode Regulator

The regulation characteristics can be improved by operating the grid from a fixed voltage divider from the choke filter output. This is:

The disadvantage of this circuit is the ripple output is slightly higher than the fixed biased case: about 200 mV for either selected output voltage. Again, this can be improved by using a larger capacitor at the grid of the tube. The load regulation of this circuit is:

Notice that this is a good improvement, and as long as the input remains constant, the output is reasonably constant. The line regulation  is slightly better than the self biased case. For the 250 volt case, with a +/-20% variation, the output changes from 198 to 305 volts and for the 200 volt case, from 158 to 245 volts. (at 50mA load).


"Improved" Fixed Biased Triode Mode Regulator

By noticing that the regulation of this supply is actually pretty good as long as the input to the regulator is constant, this suggests a "cheap" improvement. We could separately peak detect the output of the bridge rectifier (requiring an additional diode and an additional capacitor). Since the control circuit is relatively low current, and doesn't change with load, this means the detected peak will be relatively constant. This changes our circuit to:

This also has a slightly lower ripple (about 150mV for either selected output voltage) and improved regulation characteristics of:

The line regulation is about the same, though: for 250v nominal, a +/- 20% change in line changes the output from 200 to 306 volts, and for 200 volts, changes the output from 158 to 243 volts (50 mA load).


Fixed Biased Pentode Mode Regulator

A pentode can be considered as a triode from grid to screen, and the plate as a current source. Thus, if we tightly control the screen grid, the same low output resistance can be achieved (good load regulation) with an immunity to plate variations. This, in principle could produce even better performance. Since the screen current is relatively low with respect to the plate current, we could also run the screen grid from our peak detected voltage source. The circuit is:

In addition, since the screen characteristics now determine the output regulation, the ripple is reduced to about 50mV. The load regulation characteristics are:

There are two points to note here: first, the overall regulation is even better, and the voltage variation with load is almost purely resistive. This would, in principle, provide a better "sound" quality in the amplifier run from such a regulator.

The variation with line is essentially similar to all the others: I measured 197V to 306V for the 250V case, and 155V to 245V for the 200 volt case (50mA load).


"Light(er) Weight" Versions

There is still one more variation we can consider. Since the pentode plate provides a high impedance isolation, ripple on the plate lead should not be coupled to the output. So, let's get rid of the choke. Here's the circuit:

Guess what? The load regulation characteristics were essentially the same as the previous version, and the ripple increased to only 60mV. Line regulation was the same as the other circuits.


This provides the basics for simple one tube regulators. If you look at the curves again, you will notice that the effective output resistance of these regulators approaches 1/gm. This provides an indication of the performance you can expect with essentially any tube type in your circuit.

In the next part, we will add a voltage reference tube, to obtain some line as well as load regulation.

-Steve

Part 2 Simple regulators with line and load regulation.

In this section, we will discuss the addition of a voltage reference tube to our one tube voltage regulators. This allows the regulator to provide "line" as well as "load" regulation. This is perhaps the simplest combination that offers performance expected of a voltage regulator: namely, the voltage output is constant.

There are some new concepts introduced in part 2:

  • Regulation Characteristics shown as a function of line and load.
    • Load Regulation Limits
    • Line Regulation Limits
  • Minimum "dropout" requirements of the regulator tube
  • Power dissipated in the pass tube
  • Voltage reference devices.

The same basic "unregulated" power supply is used throughout this section. It is the same one used in part 1:


Voltage Reference Tubes

The idea to be conveyed here is a sort of self contained mini voltage regulator. These devices maintain a constant and fixed voltage over a reasonable range of currents. In the solid state world, these are ZENER DIODES. In the tube world, these are rare gas filled diodes with no filament. There are perhaps 3 important classes of these devices.

  1. Voltage regulator tubes. Examples are 0A2, 0A3, 0G3, VR-150. These devices provide a relatively constant voltage over a moderate range of currents, typically 5 to 30 or 40 mA. Voltage regulator tubes will typically change 2 to 4 volts over their operating range of currents, and the absolute voltage will change a few volts from device to device.
  2. Voltage reference tubes. Example 5651. These devices ore intended to be operated over a narrow range of currents and are designed to produce a constant and repeatable voltage that remains stable over time. These are usually used in feedback multiple control tube style regulators (in other words, complex regulators). They produce good references, but are not intended to be "regulators", as the operating current range is very narrow.
  3. Neon or Argon "light bulbs". Example NE2-H. These devices make relatively good low current regulators, except that they are not really characterized for this service. Voltage and regulation characteristics vary from device to device, and the "turn on" characteristic varies with ambient light intensity. Still, they do make cheap regulators, particularly if you're going to "adjust" the output voltage anyway. Be cautious: some neon lightbulbs have a built in resistor in series with the bulb, so that they can be directly connected to line. These don't regulate.

Unlike solid state zener diodes, these devices also require a higher voltage to "start" them than their operating voltage. For instance, an 0D3 regulates at 150 volts, but you may have to apply a current limited 185 volts to get it to start. Because of this characteristic, it is impractical to place a large capacitor across these devices, since in the best case, they will form a "relaxation oscillator", and in the worst case, explode if you do so. On the other hand, they are usually much quieter than the equivalent solid state zener, and have a better "temperature coefficient". [This latter means the voltage stays fairly constant as the device heats up. High voltage zeners will change quite a bit as they heat up].

Note that voltage regulator tubes are, in fact, shunt regulation devices. However, with a 5 to 40 mA current rating, these are appropriate for shunt regulators only at relatively low powers. In part 5, we will discuss using a shunt tube to form a higher power shunt regulator.

The way you use voltage regulator tubes is actually pretty easy: at nominal conditions, choose an operating current of something like 15-20mA. Then, choose a dropping resistor from the supply to the plate of the VR tube to supply this current. For instance, suppose the supply voltage was 450 volts when it produced 15mA. Suppose we want to regulate to 300 volts, using 2 0A2 VR tubes. Then, we would need to drop 15mA at 150 volts, for a resistor of 10k. If the input dropped to 400 volts, this current would decrease to 10mA. If the input increased to 500 volts, the current would increase to 20mA. All of these conditions are within rating for the 0A2. If they were not, we would need to juggle voltages and resistance values until they were. Just remember the supply must be high enough to light the tubes. In this example, at least 370 volts.


Operating "line" limits

In part 1, I mentioned how the output voltage changed with a change in line voltage. Those limits were +/- 20%. Now, that's a pretty big change. For a nominal line of 117V, 20% change is equivalent to a change from 94 volts to 140 volts (or from 188 volts to 280 volts). Probably a more "normal" change is +/- 10%, which is equivalent to a range from 105V to 129V. (or multiply by 2 for 240V mains)..

I will show a 10% variation in line in the circuits in this section, just to give you an indication of "what happens". When we get to a "real" example, we will probably design for +/-10% changes as well.


A simple triode regulator with line and load regulation.

This looks a lot like the "improved" triode regulator we discussed in part 1. But there is one important difference. In this circuit, I have added one resistor and 2 series connected voltage regulator tubes, to provide a 300 volt reference. This reference will stay about 300 volts as long as I can provide enough voltage to "light" the regulator tubes. The voltage divider setting the control grid voltage is thus made constant with line variation.

The regulation characteristics of this circuit are:

There are several things to note on this regulation picture. First, note that in addition to plotting both 200V and 250V "set" points on the same picture, I have also shown the line voltage variation. Also notice that the vertical scale has been expanded. This allows you to get an instant picture of all the conditions you will likely encounter, and the improved regulation allows an expanded scale to see more detail.

This second point is important. For the previous "improved triode" regulator, at 50mA draw, the voltage varied from 226V to 250V to 276V as the line changed from -10% to nominal to +10%. With this circuit, the change at 50mA is 239V to 254 volts! Because the voltage on the plate does vary, even maintaining a constant grid voltage allows the output voltage to vary somewhat.

The ripple from this supply was about 100mV. This improvement is the result of the additional filtering action of the regulator tubes. There is only a couple mV of ripple on the grid!

There is one additional item to note with this circuit. Notice that the "low line" 250v curve indicates less regulation at higher load currents. This is due to the tube "saturating". There is not enough plate voltage to maintain good regulation at the cathode. This is called regulator dropout. I'll discuss this in more detail later.


A Simple Pentode Regulator with Line and Load Regulation.

Similar to the pentode regulator we discussed in part 1, the voltage regulator tubes can be added to that circuit as:

Again, one resistor and the 2 VR tubes have been added. Note, however, that this time I'm also running the screen voltage from the regulated supply. Since there is some added screen current, I've lowered the dropping resistor from 8k to 5k, in order to supply this added current.

Here is the regulation characteristics that can be achieved:

This is actually quite good as a regulator. Notice that the stiff screen source and stiff control grid source allow essentially "perfect" regulation. [meaning that the output resistance is 1/gm. The curves indicate an output resistance of about 70 ohms!]. Also notice that the line regulation is also quite good,;it's about 0.5% for +/-10% line voltage change.

The output ripple was about 20mV, due to the plate isolating the output ripple from the LC filter, and the VR tubes providing low ripple to the screen and control grids.

There is one additional benefit of the pentode in this application. The tube can maintain regulation when the plate voltage is lower than the screen voltage. Notice the low line 250V case is much better (but not perfect... in fact, the imperfection here is not due to the plate "saturating", but rather the increase in screen current starting to pull the VR tubes out of regulation range.)  It is the lower available voltage drop of the pentode, along with its constant current characteristics that provide the dramatic improvement in performance in this simple circuit.

Again note that since there is no feedback conditions (other than cathode follower degeneration), there is no major "frequency response" issues with this regulator. It will "sound" pretty good. I'll introduce the concept of plotting the output impedance vs frequency when we get into regulators with defined feedback loops later in this series.

Note that this circuit could be turned into a variable output power supply. Simply replace R1 and R2 with a 100k to 500k pot. Do place about 10k in series with the point connecting to the VR tubes though, so that when the wiper is turned so that the capacitor on the grid connects towards the VR tubes, there is still 10k resistance isolating the VR tubes from the capacitor. Available output voltage effective control range is from about 50 volts to about 275 volts. We will discuss the available current below.


Supply Limits and Tube dissipation

The realistic limits on the regulated power supply are controlled not only by the power supply (transformer, rectifier, filter inductor) but also by the tube. The tube provides a limit in current and power. Lets initially assume the power supply components do not limit the circuit (they are big enough). Then, what limits the output voltage and current?

In the above circuit, the screen voltage is held at 300 volts. Thus for cathode (output) voltages approaching this value, there would not be enough conduction. A practical limit in this instance is probably 275 volts. Also a practical limit is also probably about 30 volts (anode to cathode) drop. For the rectifier and filter combination, this means the filter output would have to be 275+30=305 volts. This occurs at about 35 mA. At this point the tube is dissipating 30v*35mA= about 1 watt. At 275 volts, the maximum output current would be about 35 mA. If we content ourselves with 250V maximum, the screen to cathode voltage is 50 volts, so the minimum anode to cathode voltage might be about 25 volts. This occurs at about 120 mA, so 120 mA is the maximum available current at 250 volts. (The tube is dissipating 25v*120 mA which is less than a watt).

If we consider -10% line conditions, 275 volt output would be limited to about 20 mA, due to the filter output voltage (305V at 20 mA), and 250V output current would be limited to about 50 mA. Notice that the curves above confirm this analysis: at 250 volts, above 50 mA, the output voltage starts to drop. I should add, though that I'm ignoring screen current, which is not *quite* fair. In fact, the screen current from the "higher voltage" supply is attempting to provide about a third of the output current at 50 mA (i.e., 16 mA) under low line high voltage conditions, and this is starting to pull the VR tubes out of regulation.

At the low voltage end of our hypothetical variable supply, power dissipation and./or maximum tube current will limit the available output. Let's consider the case of 50 volts output. (Potentiometer turned towards the ground end). In this case, we have 250 volts screen to cathode... we can get lots of conduction. The plate voltage will be relatively high. At high line and 200mA, the filter output is about 290 volts. Thus, 240 volts and 200mA flows in the tube. (48 watts). This exceeds the 35 watt tube rating. In fact, working thru the math implies we can draw about 140 mA from the supply at 50 volts. At 100 volts, we could draw about 180 mA. At 150 volts we could draw about 240 mA. At 200 volts, we could draw almost 500mA. (At high line, the filter output produces about 270 volts at a 450 to 500 mA load. The output will have dropped to about 195 volts for 75 volts plate to cathode. For 35 watt dissipation, this is 466 mA) However, at 200 volt output and LOW line, the filter output will be about 220 volts at 400 mA, so about 400 mA is the limit at 200 volts.

Notice that in some of the conditions above, the limit was high line, while in other conditions, it was low line.


In the next part, we will talk about error amplifier tubes and demonstrate their effect on power supply regulator performance.

-Steve

 The Error Amplifier

This part describes the use of an Error Amplifier in the voltage regulator to improve the regulation characteristics that can be obtained. It uses the principle of feedback to attain these goals. Since it is a feedback process, there will be some advantages and some disadvantages to this type of voltage regulator. The positive side is better regulation, lower output impedance and less "hum" from the regulator. The negative side is the potential for oscillation if the feedback loop becomes unstable. This could happen with odd power supply loads. Also, the output impedance will change with frequency, as the feedback loop is not infinite in bandwidth. These negative effects sometimes outweigh the positives, but usually the positive benefits "win".


In its broadest form, the error amplifier compares a fraction of the output voltage with some fixed reference, and uses the derived "error" voltage to control the series pass tube. Thus, the output impedance can be made much lower than the 1/gm we were able to obtain with simple "cathode follower" regulators discussed in parts 1 and 2. It is actually pretty easy to predict the improvement that you will achieve with this form of regulator. There are two additional factors: the voltage divider, and the gain of the error amplifier stage. Suppose we choose to use a 100 volt "reference". (You remember how... we use a VR gas tube reference device to establish this voltage.) Suppose further we want a "regulated" 300 volts output. In this case we would divide the output down from 300 to 100 volts, and apply it to the grid of the error amplifier. The reference could sit at the same tubes cathode, providing a low DC resistance from the cathode to ground. The voltage divider produces a loss of 3:1 (in this case). Suppose we use a small signal pentode or cascoded triodes for the error amplifier. We are likely to get a "gain" in the error amplifier of perhaps 150. Thus, the overall "gain" in the feedback loop is about 50. (150/3=50). The effective output resistance for DC will DECREASE by this factor. This means that our 70 ohm output impedance (from part 2) could be reduced by this factor of 50 to 1.4 ohms. This is equivalent to to a regulation of (no load to 100 mA, for instance) 0.14 volts! In other words, if the no load voltage were 300.00 volts, with a load of 100mA, the voltage would drop to 299.86 volts!

For AC purposes (hum reduction), we can "bypass" the voltage divider with a capacitor, making the AC reduction a full 150. This means that we could reduce the 20 mV ripple we achieved in part 2 to 134 MICROvolts.

OK, so what's it look like?


There is a common circuit that has been around for (it seems like) about 10,000 years ;-)

This is a common circuit, that shows up with different part values in many places. I have altered the values to produce about 200 volts output. The output voltage is adjustable over a limited range by altering R1 and R2. In addition, the 0.1u capacitor from the output to the error amplifier grid has two purposes. First, it provides more error injection of the hum components, as mentioned above, to provide better ripple rejection. The second purpose is to aid in stability of the feedback circuit. Sometimes you will also see a small capacitor across the 0B2 VR tube. Its purpose is to remove any high frequency noise generated by the VR tube. DO NOT EVER use a large capacitor across these tubes. When used, a 0.01uF or so is about as large as you should use, although I've seen this same circuit with a 0.1uF across the VR tube.

This circuit provides pretty good regulation.

Note that since this circuit provides even better regulation than those of part 2, I have again changed the scale! The ripple measured <1mV at 100mA output. It also has, as you might note in the above graph, the ability to SINK current as well as source current, which keeps any potential transients coming from an output stage, for instance, from causing the regulator to "un-regulate".

There are a couple deficiencies with this circuit, though. By placing the reference on the error amplifier cathode, there is now a definite minimum voltage that can be achieved (a disadvantage if you wanted a universally adjustable regulated supply). The 105 volts on the cathode implies that the plate of the error amplifier can go down to about 130 volts. So the grid of the series pass tube can never go lower than 130 volts. This makes the practical lower limit of the output somewhere about 155 volts (allowing 25 volts to "cutoff" the series pass tube, for low currents). The upper limit is the supply voltage produced, the maximum desired current, and the voltage drop across the series pass tube. With the values shown, about 220 volts at 100mA is about the limit. Since the cathode of the error amplifier is elevated, both the series pass tube and the error amplifier tube need to have separate heater windings, to keep the heater to cathode voltage in range, particularly if a wide range variable supply is sought.

Also, the additional capacitors in the circuit, along with the feedback style regulation cause the output impedance to vary with frequency. If you take the average slope of the above regulation curve, the effective "DC" output resistance is about 1 to 1.5 ohms. Here's the AC impedance vs frequency:


Some of the deficiencies of this circuit are relatively easy to fix, however. We could provide a -150 volt reference (which could then be used for bias purposes), and instead of grounding the bottom of R2, we could connect it to -150 volts. We could then ground the cathode of the error amplifier. This actually allows better regulation, since the error amplifier cathode impedance is even lower, and the voltage division ratio turns out to be more favorable.

This style regulator looks line:

This style regulator had virtually identical regulation performance to the one above, and about the same ripple characteristics. Basically the lower cathode impedance at the error amplifier (ground) offset the slightly different voltage divider action of the grid. I also change the 0.1uF capacitor to 1nF. This provides a more uniform output impedance over the audio frequency range (about 0.4 ohms). Since the cathode of the error amplifier is grounded, there is no problem finding a source for the heater voltage. Of course, you need to provide a negative supply for the 0A2, but this is realistically probably necessary for bias purposes anyway. The screen dropping resistors change, but the current is about the same, so this supply also had the same "sink" current capability.

The other advantage of this circuit is the ability to provide lower output voltages if needed, so it is much more flexible when considering a regulator providing a variable output for "experimental" purposes.


Error Amplifier Tube Characteristics

What makes a good error amplifier tube? From the above discussion, these characteristics become fairly obvious. For higher voltage applications, the error amplifier "sees" essentially the entire output voltage, so a high voltage device is needed. The current and power ratings are not usually too important; note that in the above circuits, there is not much power nor current going THROUGH the error amplifier. A high gain is desirable. This is why the typical error amplifier tube is a "high voltage" pentode. This is also why I used a 6BM8 as the above example. The high gain and high voltage capabilities are quite nice. (It also has an additional triode section which we will use in part 4 of this series).

A 6SN7 would work, for those who MUST use a triode, but you would probably need to place it in a cascode configuration in order to achieve sufficient gain. The limitation would then be the minimum upper plate to lower cathode voltage differential limiting the MINIMUM voltage.


A Practical Example

It's time to put the fundamentals we have been building in this series to a practical example. Lets consider building a "universal" regulator that is capable of providing a regulated output in the range of 50 to 450 volts, with a few hundred mA current range over most of the supply voltage range. We want the output resistance to be in the ballpark of one ohm (0.2V change for 200mA current change), and we want the power supply output ripple to be in the range of 1mV or less. We will also provide a fixed negative supply of 150 volts for bias.

For the main power transformer, I will consider the BFT-1, with additional ancillary transformers as necessary to cover the voltage range and other requirements as needed. (Caution I don't own a BFT-1, so I'm using transformers I have that I believe should perform the same, but there's still the off chance I messed up). I will break the schematic into two portions; the first is the transformers, rectifiers and filters, the second portion is the regulation circuit.

Here's the transformers, rectifiers and filters:

And here is the regulator:

The range of adjustment with this supply is from about 50volts to about 450 volts at nominal line and 50 volts to 410 volts with a 10% low line. (Due to the dropout voltage of the regulator.) The low voltage limit is due to the screen voltage placed on the error amplifier (25 volts at 50 volts output).

I have provided a number of things with this regulator:

  1. An isolated 6.3VAC 9.6A AC heater supply for your experimental amp.
  2. An isolated 6.3VDC at 2A max heater supply. This is filtered but not regulated.
  3. A -150V at about 15mA max bias source. This is regulated and has about 200 microvolts of ripple.
  4. A variable regulated supply with about 0.7 ohm output impedance, relatively flat across the audio band. This supply is variable from about 50V to about 450 volts. Hum level is less than 1mV at any valid output condition. (See table below). Typical hum level is about 250 microvolts, even with no choke in the supply. I will caution you, though, that to achieve this hum level, you have to be careful about your grounding. The high ripple currents from the filter caps should not be "in series with" the error amplifier in finding a ground. The schematic sort of suggests the order of grounding things.
  5. There is some current sinking ability with this supply as discussed above.

The maximum current available from the supply is a function of the output voltage you set. At high output voltages, the output current is limited by the dropout voltage (I measure about 30 volts at 150mA), and at low output voltages by the power dissipation of the tube. Reasonable limits are:

Regulator current capability
Output Voltage Maximum Current
50 50mA
100 80mA
150 100mA
200 120mA
250 150mA
300 210mA
350 350mA
400 250mA
450 125mA

Here is typical regulation characteristics (measured at 350 volts output).

Notice that the performance is just slightly better than the previous example: the DC output resistance of this supply is only about 0.7 ohm!


An Interesting Tweak

If you connect a resistor of about 850k from the EL509 plate to the 6MB8 screen grid, then as the load current increases (causing the plate voltage of the EL509 to decrease), then you add additional compensation. For DC conditions, if you adjust the value of this resistor, you can achieve essentially perfect compensation: the load voltage REMAINS THE SAME regardless of the load! However, this is true of static conditions only. Here is one case where the DC output resistance is zero, but the AC output impedance is not zero. Looks fine on "specs", but, to my ears anyway, doesn't sound good, and looking at the output of the supply, feeding an amplifier shows some less-than-desireable signal. I mention this as a caution against blindly following specs, without really understanding the meaning behind them.


Can we do even better than this?

-Steve

 Part 4 - Multiple Control Tube Voltage Regulators

There are a couple different ways in which multiple control tubes can be used in voltage regulator applications. In this part, we will discuss a couple of the more common uses.


One reason to use multiple control tubes is to achieve more open loop gain in the regulator so that the voltage regulation is even better in the closed loop case. The problem with this approach is that as the open loop gain increases, it becomes more difficult to keep the "closed loop" regulator stable, particularly with different type loads. Characteristic of this type of regulator will be multiple capacitors that are used to tailor the regulator "frequency response" to maintain stability. Here is a circuit typical of this type regulator:

In this circuit, the output changes are mirrored at control tube #1's cathode, due to the coupling via the VR tube. These changes are amplified by control tube #2, used to control the pass tube. The ripple rejection of this kind of circuit is usually pretty good, and the DC regulation not too bad. One can typically get about 0.5 ohm output resistance (0.1 volt change for a load from 0 to 200 mA.) Because of the additional capacitors in the circuit, the AC and DC characteristics of this kind of regulator are usually quite different. Because of the multiple control tubes, and multiple VR tubes, it is usually difficult to get a wide range of output adjustment from this kind of voltage regulator.

One could add a 3rd or 4th control tube, but this is uncommon, since it becomes increasingly difficult to keep the resulting regulator circuit "stable".

Cancellation Circuits

The other common use of a second control tube is to provide a cancellation mechanism for AC ripple, or DC regulation characteristics. I mentioned briefly in part 3 that an additional resistor could be added to the example regulator to provide essentially perfect compensation. By using a second control tube, it is possible to achieve this compensation without many of the disadvantages of attempting to do it with one control tube. Some of you might have noticed that the 6BM8 tube we used as the control tube in the regulator of part 3 is a dual tube; it also contains a high mu triode section that we did not use.

We can, however, modify the circuit of the regulator described in part 3 to use this second section:

In this modification, some of the ripple as well as some of the line-change-with-load is fed into the grid of the otherwise unused part of the 6BM8. This is not a feedback mechanism, rather it is a "feed forward" style of compensation. It is usually possible to adjust the DC pot (R18) for either perfect line or perfect load regulation, but not both simultaneously. It is also possible to adjust the AC pot for perfect ripple cancellation or very low AC output impedance, but not both simultaneously. Here is the regulation characteristics I was able to achieve by adding this circuit.

Note that again, I had to change the scale of the vertical axis. Each box represents 2 millivolts!

The best ripple performance I was able to achieve with the values shown was 15 microvolts of output ripple, at any one output load, and about 25 microvolts ripple over the range of 0 to 350mA load. The DC characteristics I've shown above are just about at the limits of what I can measure with any degree of repeatability!

Sound? I couldn't make a definitive statement whether the part 3 regulator or the tweak shown above sounded better. I could convince myself that I liked either one better, but I probably would not be able to tell the difference in an "honest" A-B test.


This pretty much covers series pass regulators. In the final part of this series, I'll cover shunt regulators.

-Steve

 Shunt Regulators

Shunt regulators of the simple type are often used as part of series regulators. We used them in parts 1-4 of this series as VR tube references. They are the classic "zener" regulators. However, these have a limited output capability. The final part of this tube regulator series investigates not only the classic VR tube references, but also single tube shunt regulators and feedback multiple tube shunt regulators. This appears to be not widely applied. I could not find much in the way of references to tube controlled shunt regulators, except for very high voltage and low current applications. (The old tube based color TV's used a shunt regulator to maintain the picture tube "anode" voltage; typically using a 6BK4 as the shunt regulator element. Ditto for oscilloscope HV supplies).

Shunt regulators have a different set of constraints than series regulators. The highest power dissipated in the regulating element occurs at zero external load, and the maximum available load current is thus determined by how much current you care to put through the shunting element. Their output voltage -vs- load characteristics tend to be very different as well. When the maximum output current is exceeded, rather than "following the unregulated supply" down in voltage with increasing current, the output impedance is very much higher: being the resistance of the current limiting element is series with the shunt. (More about this later). Also, shunt regulators are capable of sinking substantial current as well as sourcing it. This is useful if your load can kick current back into the supply (inductive loads can do this). The penalty for this characteristic is the current goes through the shunting device, so this must be taken into account when "sizing" components.

Shunt regulators tend to be much less efficient (get hotter) for a given output power. You can picture this relatively easily: in both cases, the required output current flows through the regulating device (either at no load or at full load). However, the voltage drop across the series pass device is only the input-output differential, whereas in a shunt regulator, the entire output voltage is across the shunting device. Because of these constraints, it is much more difficult to design a variable shunt regulator with a wide range of output voltage and wide range of output currents.


Power Supply used for all the examples in this part.

I used the simple rectifier-filter that was designed for the "example" in part 3; namely, the BFT1 with a capacitor -only- filter. This was the input to ALL the regulators shown in this part. The regulators we will examine are simple VR tube types, single control tube shunt regulators, and multiple tube shunt regulators. Output voltage picked for these examples is about 275 volts.


Simple VR Tube Shunt Regulators

This is probably the simplest form of shunt regulator. Lets us an oA2 and an 0B2 in series. This should produce about 265 volts. Since the power source produces about 490 volts when loaded with 20 or so mA, I'll use a 10k resistor to drop the voltage to the 2 series VR tubes, passing slightly more than 20mA thru the tubes. It looks like this:

One thing to note about the VR tubes. They generally have a few "extra" connections in their "basing". In the example shown, this is used to remove the output voltage if the 0A2 is removed. Likewise, I could have placed the output ground  in series with pin 7 of the 0B2, to remove the current flow path if the 0B2 were removed. This is sometimes used for protection.

The regulation characteristics are:

Notice that the effective output resistance is about 66 ohms. With only the C filter, the output ripple was about 10 mV. This is good ripple attenuation, but only because only 20mA can be drawn from the supply. With about 20 mA flowing thru the VR tubes at no load, you can also source about another 15mA FROM the load into this regulator and still maintain operation within all tube ratings.


Single Tube Shunt Regulator

It is possible to use a single control tube, like we did in part 2 of this series to achieve more output current and a bit better regulation characteristics. A typical circuit looks like this:

In this circuit, the top 0A2 drops the voltage to the screen of the 6KG6/EL509, and the second 0A2 further drops the voltage to the control grid. The 0A2 current is returned to a presumed -150 volt source. This establishes about -20 to -25 volts on the control grid. At no load, all the current flows through the EL509. As output current is drawn, the voltage would tend to drop, lowering (making more negative) the control grid voltage on the EL509, so it draws less current, compensating for the increase in load current. If current is SOURCED into the output terminals, the voltage would tend to rise, causing the control grid voltage to rise, causing the EL509 to conduct more heavily, thus "regulating" the output. With 40 watts available (plate + screen dissipation), and for about 275 volts output, about 150 mA can be drawn through the tube. An additional 5 mA can be provided due to the current through the 0A2s. Therefore, we can use a 1.2k dropping resistor from the source. In this example, I used a 1.5k resistor. The regulation characteristics are:

Note that the effective output resistance is about 40 ohms. Also note that the curves are notably different form the simple series regulator as I described above. (Specifically note that with series regulators, the regulation is "better" at high loads, and gets worse at low or negative loads. With the shunt regulator, the regulation is better at low or negative loads, and gets worse at high loads). Technically, since the tube is dissipating its rated power at zero load, "sourcing" current would over-dissipate the tube. For dynamic loads this is not a problem, since the power averaged over a load "cycle" would still be within limits.

The ripple output from this regulator was about 50mV. Even though the "regulation" is slightly better than the simple regulator shown above, the added output current caused the ripple on the output to increase. This could, of course, be improved by using a C-L-C filter before the regulator.


Shunt Regulators with multiple control tubes

It is possible to use multiple control tubes to provide more effective shunt regulation. This also makes it practical to achieve some level of output voltage control too. One such implementation that is usually "trouble free" to bet operating is:

Note that I have also "presumed" a -150v regulated source is available. I've also shown the "quiescent" voltages for this supply to make it more obvious how it operates. The added tube is a differential amplifier that amplifies the "error", sensed from the bottom of the 0A2 and referenced to the -150V source. Some level of output voltage adjustment can be achieved by altering the 33k/39k voltage divider ratio. Since I am placing about 13 mA thru the VR tube, and the 12AX7 draws an additional 2 mA, I have lowered the series resistor to 1k. (So I can actually get slightly more current from this supply without exceeding the EL509 tube limits.

Here's the regulation characteristics achieved:

Note that the output resistance is about 2 ohms, which is not too bad! The output ripple from this regulator was about 4mV at a 100mA load.

One other thing to note is the use of the multiple connections to the 0A2. If you remove the 0A2, the output is disconnected, and the screen voltage to the EL509 is removed, so that the EL509 is not over-dissipated, and the load is not "over-voltaged".


I hope you have enjoyed this series on voltage regulators.

-Steve

 

Of Loadlines, Power Output and Distortion

Purpose

In this article we will explore establishing a loadline on a triode vacuum tube (valve), determine the power available from this loadline, and the distortion predicted from this loadline. This will allow the reader a step by step way to calculate the parameters for his design. I have chosen to "invent" a tube for illustrative purposes. My invention is "spec'd" as a triode with the following parameters:

For this article, I will stick with a single ended transformer loaded Class A amplifier, as one of the simplest vehicles that can convey this knowledge. Most of the information is appropriate for push-pull operation, once a composite load line is established (may do that in another article).


Establishing a Load Line

  1. Get the tube characteristics for your design (shown above for my example). If not already on the characteristics, plot the maximum power dissipation curve (voltage * current = spec'd power).
  2. Select a reasonable quiescent operating point for your design. Often this is described in the manufacturers typical operation. For instance, a reasonable starting point for my example would be 300 volts and 60 mA. (This is the orange "dot" on the characteristics above.) Whatever you pick MUST BE within the tube's ratings. For this example, I'll choose to ignore the manufacturers recommendation, and use 300 volts and 65 mA. Put a dot on the tube characteristics at this point. (I am expecting that the "recommended point" is somewhere near optimum, but I want to run the tube a little "hotter", perhaps gaining more power output, lower distortion, or both.)
  3. Now start drawing a potential load line. This is a straight line whose "slope" is the primary "impedance" of the transformer. How do you do this? Pick a point somewhere about DOUBLE the quiescent voltage and zero plate (anode) current. Draw a line (called "loadline" in the curve below) thru these 2 points extending to the 0 plate volt axis. Did your loadline go above the max power curve you drew? Not good. Go back, pick a different zero current point. Is the voltage at no current more than twice the "rated" voltage? Not good. Go back and pick another point. Within spec? Good. Now determine the impedance. Z=(maxvoltage-quiescent voltage)/quiescent current. Alternately, you can pick an impedance to suit available plate transformers. In this case, if you go outside the safe boundaries, you must choose a new quiescent point. The curve below chooses a 4k load line.
  4. Having now chosen a tentative loadline, we need the following information from it......
    1. The required grid bias (you'll need to supply the grid with this value of DC voltage).
    2. The quiescent voltage (called Vq)
    3. Voltage where the grid bias voltage is zero (called Va).
    4. Voltage where the grid bias is double the quiescent value (called Ve). Note... if the plate current reaches zero before you get to this bias, you've got a bad operating point, go back to step 3 and try again.
    5. Current at 0 volt grid bias (called Ia). If this current is above the maximum rated tube current, go back to step 3 and try again. Hint: If its off the scale, chances are the current is too high.
    6. Current at half the quiescent bias voltage. (Called Ib.)
    7. Quiescent Current.(Called Ic.)
    8. Current at 1.5 times the quiescent bias voltage (Called Id).
    9. Current at twice quiescent bias voltage. (Called Ie).

Note that these are all plotted out on the following set of tube curves for our example.

For our example, the data is as follows:

Va= 99V, Vq=300V, Ve=465V, Ia=115mA, Ib=89mA, Ic=65mA, Id=42mA, Ie=23mA. In addition, 40V p-p are required from the driver stage, and a 4k plate transformer is required.


Rectification Effects

In any device with even order distortion (not just second harmonic as sometimes stated), the average current of a Class A stage will change depending on the signal level. If the distortion is relatively low, the effect is relatively unimportant, if the distortion is high, this effect becomes increasingly important. (It is also more important for self bias than fixed bias, as the bias voltage is a function of the average current).

This effect modifies all tube operating characteristics. The degree can be seen by comparing the quiescent current (65 mA in our example) with the average current as taken from the "extremes": that is, (Ia+Ie)/2. In our example that value calculates to 69mA (note that is not substantially different). The effect requires an iterative plot of the load line on the tube characteristics. The "worst case" is shown in "pink" in the constructed load lines above. The real case will be between the two extremes. Notice that the "curve" includes some portion above the maximum dissipation. In this case, the power dissipated in the tube is (300*0.069 - power output), or in this case 16.5 watts. This is still LOWER than the 20 watt max dissipation case, so there's no problem. Note the quiescent dissipation is still 300*0.65, or in this case 19.5 watts.

The remainder of this document ignores this effect: it is important in high distortion cases only, but high distortion cases are not of much interest.


Power Output

Po = (Ve-Va)*(Ve-Va)/(8*loadimpedance).

In our example this is (465-99)*(465-99)/(8*4000)= 4.18 watts. This number is an approximation in that it assumes low distortion.

Second Harmonic Distortion

HD2(%) = 75*(Ia + Ie - 2*Ic)/(Ia + Ib - Id - Ie)

In our example this is 75*(115 + 23 - 2*65)/(115 + 89 - 42 - 23) = 4.3%

Third Harmonic Distortion

HD3(%) = 50*(Ia - (2*Ib) + (2*Id) - Ie)/(Ia + Ib - Id - Ie)

In our example this is 50*(115 - 178 + 84 - 23)/(115 + 89 - 42 - 23) = -0.72%

Notice the minus sign. This indicates that the harmonic content subtracts from the fundamental (flattening it) when the fundamental is at its crest. This *usually* happens on third harmonic distortion in tubes.

Fourth Harmonic Distortion

HD4(%) = 25*(Ia - (4*Ib) + (6*Ic) - (4*Id) + Ie)/(Ia + Ib - Id - Ie)

In our example this is 25*(115 - 356 +390 -168 + 23)/(115 + 89 -42 - 23) = 0.72%


Discussion

If you don't get numbers you like from the above, you are free to choose different operating points and loads to alter the characteristics of the stage. This may alter the distortion, power output or both. You may freely experiment on paper until you get a "design" you like. You can then correlate the results with your own listening tests. Over time, you'll find there is a correlation between what you predict, and what you hear. Then, you will be able to come up with new amplifiers more rapidly, as you will be able to zero in on what you like by doing a few "paper designs" up front, before committing to a particular tube, transformer, or power supply.

This procedure also works for small signal tubes for driver applications, preamps, etc.

Steve

 

Part 2 - Effective Plate Resistance

One of the questions you may hear asked is how to pick an adequate load line for your design. You will hear talk about picking it with respect to the plate resistance of the tube in question.:"Pick the load resistance equal to the plate resistance to maximize power.", "Pick the load resistance equal to twice the plate resistance to maximize power.", "Pick the load resistance higher to minimize distortion.", "If you have a fixed input, the maximum power is ...".

None of that is much help. The biggest problem you have facing you is the load line. This is usually established by the characteristics of the tubes you choose, and available components (output transformers in particular). The bias, high voltage (B+ or HT voltage), and drive are usually determined AFTER you establish an output stage characteristic. Then, if one of those secondary parameters is unrealistic, you "iterate" the output stage characteristics.

Plate resistance is not constant. It varies widely over the operating load line. We will continue to use the same example as previously... my "invented" triode. In the figure below, I have plotted the tangent line to the operating characteristics we used in the previous article at the same 5 points on the load line we used to calculate distortion.

Notice that in this case, the plate resistance varies from 580 ohms to 2020 ohms. So, if you used the 2x Rp "rule of thumb" (or was that rule of middle finger?), which value do you pick? The 1000 ohms specified by the manufacturer, the 580 ohm minimum, the 1050 ohms at "quiescent", the 2020 ohm maximum, or yet another number?

In our operating example the load line varies from about 2:1 Rl/Rp (at minimum current) to about 7:1 Rl/Rp (at maximum current). AND, this assumes a constant speaker impedance. In reality, most speakers vary an additional 2:1 to 10:1 over their operating range, further complicating things.

So, what do you do? You pick an operating point as described in part 1; one that is within the tubes ratings and produces a reasonably wide operating range. If it fills your needs, you're done. If it doesn't, start over again. Moral of story: don't be particularly concerned with the load lines relationship to the plate resistance. As we've just seen, the plate resistance varies.


Effect of Variable Plate Resistance

You might wonder what the effect of the variable resistance is. This will change the damping on the speaker over the signal range. For instance, let's consider a band limited square wave (one with rounded edges that "fits" within the bandwidth of our amplifier). The speaker might "ring" on the "edges" of this waveform. That's usually why higher "damping factors" are usually considered desireable. In this case, the ringing at the edge associated with the high current point will be damped MORE than the ringing at the edge associated with the low current point. This is a case where the amplifier cannot correct for a speaker defect.

There is another, in some sense, worse effect. Let's further "modify" our hypothetical source by placing a smaller, high frequency sine wave component on top of the square wave. I'll stipulate that this signal's frequency happens to be at a relatively low speaker impedance point. Now what happens. The portion of the sine wave riding on the high current part (lower impedance) will have higher "gain" than that portion riding on the lower current part of the loadline. This is an insidious form of intermodulation distortion that is usually not very musically pleasing. You can somewhat minimize this effect during the establishment of your load line. If you followed the procedure I outlined in part one, go back and re-plot the loadline using HALF the resistance, then TWICE the resistance. Don't be particularly concerned about whether the newly plotted loadlines remain inside the power dissipation characteristics. (Why THAT is, is outside the scope of this particular article). Now calculate the distortion FOR APPROXIMATELY THE SAME power output on these 3 loadlines. Continuing with our example, this is shown below. The original loadline (4k) is green, the 8k loadline is orange, and the 2k loadline is magenta. I've shown the current and voltage points needed by black crosses on the curve.

Notice that the input signal levels for the highest impedance is twice that of the lowest impedance, and the original curve is intermediate between these. The points to pick are limited in two ways: The signal for the highest impedance will usually cause the limit at "0" bias. This fixes the max signal. Sometimes the signal at the lowest impedance reaches cutoff. If that happens, choose a maximum level to remove this restriction, and re-derive the points for the other impedances. Caution: If this happens it is usually an indication that the load line you originally picked is not close to optimum).

Results:

Remember the formulas we used:

Po=(Ve-Va)*(Ve-Va)/(8*RL)

HD2=75*(Ia+Ie-(2*Ic))/(Ia+Ib-Id-Ie)

HD3=50*(Ia-(2*Ib)+(2*Id)-Ie)/(Ia+Ib-Id-Ie)

HD4=25*(Ia-(4*Ib)+(6*Ic)-(4*Id)+Ie)/(Ia+Ib-Id-Ie)

The data points are:

8k:  Va=86, Ve=500, Ia=92, Ib=79, Ic=65, Id=53, Ie=40.

4k:  Va=150, Ve=427, Ia=101, Ib=82, Ic=65, Id=49, Ie=32

2k:  Va=214, Ve=369, Ia=107, Ib=85, Ic=65, Id=46, Ie=30

These lead to the following results:

Load Z Pwr Out HD2(%) HD3(%) HD4(%)
2k 1.5 4.5 -0.4 0.6
4k 2.4 2.2 1.4 -0.3
8k 2.6 1.9 0 -1.9

You can see that the distortion at the lowest impedance is highest, but is comparable at all 3 impedances. If these were wildly different (say 1.5% to 10%), this would be an indication that your amp is likely to sound less than desireable, and it might be wise to choose different operating points.

What else can you do? Feedback is one possibility. I generally have good luck with some slight local feedback from the output stage plate to the driver plate. This does not suffer from a lot of the global feedback effects that shun people away from feedback in general. You are free to go back and start a different set of load lines, to see if you can improve things.

In a later installment, we will see how push-pull operation affects the effective plate resistance.


One additional Clarification

In the example we have been using, the quiescent voltage is 300 volts. In building the power supply for this amp, don't forget to include the voltage drop in the output transformer. If this were, for instance, a 110 ohm winding, the power supply needed would be 307 volts. If this were, in addition, a "self biased circuit" (i.e., bias derived from a cathode resistor), the 20 volts bias would also needed to be included, bringing the power supply required to 327 volts.

Steve

 

 

Purpose

In this installment we will discuss how to create a composite load line. This is "necessary" to establish the load line for a push pull amplifier, regardless of whether it is going to be Class A or Class AB. We will continue to use the same "invented" tube for this discussion. I will also use Class A, and the same "operating points" we used in the single ended examples. The data points used are consistent with the other 2 parts of this series, you should choose the operating point for your particular case as described earlier.

Building a Composite Load Line

This is again an iterative process, perhaps moreso than the single ended case as it's a bit more involved. Here's a step by step procedure to create a set of "composite curves" and load line.

1. Establish your intended operating point from the single ended curves. For this example, we will choose 300 volts and 65 mA quiescent current (-20 volt bias). This is the same operating point chosen for the single ended case. Note: For Class AB, you may choose a lower idling current point, which will allow a lower impedance loadline (and higher power output) without exceeding maximum allowable power dissipation. For instance, choosing a -25 or -30 volt quiescent point would allow you more "room" to increase tube current. To stay within Class A operation, at least some current must flow in each tube at all bias points, otherwise, you will be operating in Class AB.

2. Since the plate voltage increases on one plate and decreases on the other plate, we must have a way of representing this. The usual method is to take another set of the same tube curves, turn them around (so the maximum current is "down" and highest plate voltage is to the "left", and position them so that the quiescent voltages line up vertically, and the zero plate current lines touch each other as shown below:

3. Now, for the chosen bias point, create a new line (labeled "composite" curve) as follows (we are going to look at ONLY the -20 volt bias lines, one on the "upper" part of the graph, one on the "lower" part of the graph:

  1. At the quiescent point (300V), subtract the currents (65-65=0 mA). This is the magenta points on the curves. Draw the "dot" at 300V, 0 mA on the composite curves.
  2. Step to the next convenient voltage, (the dark magenta ones) note the currents in the upper and lower parts of the graph, subtract the two, and plot this point on the composites.
  3. The next points are the ones shown in blue. Continue until you have the entire curve "dotted" for the -20 volt lines. The new "plate curve" for -20 volts is shown in the graph above.

4. You have now established one "line" of the composite curves. This is the line of varying the plate voltage symmetrically about the quiescent point while maintaining a constant grid voltage. (classic plate resistance line).

5.  Now consider another set of lines for your graph. This will be the -25 volt line for the "upper" set and the -15 volt line from the lower set. (We have put a 5 volt signal into the push pull stage). Again, for each plate voltage, subtract the two currents (for instance, 310V -25V bias upper and 290 volt -15V lower) and plot the dot. Continue until you have the -25/-15V line filled in. Note that if your quiescent grid voltage was, for instance, -30 volts, the second line you would add would use the -35/-25V lines instead of -25/-15V lins.

6.  Next do the same for the -30/-10, the -35/-5 and -40/0 volt lines. Then do the same for the -15/-25, the -10/-30, the -5/-35 and the 0/-40 volt lines. This completes the composite curves. The last 4 curves should be mirror images of the previous 4 you filled in. Noticing this saves you some time.

7. The last thing to do is fill in the maximum power dissipation curve. There are now going to be 2 of them, one for the one tube, one for the other. These will be symmetrical about the quiescent point. It is usually necessary to graph only half of the curve. The following is for the "upper" tube. Here's how to do it:

  1. Get the current at the quiescent point. (In our case that's -20 volts, 300V, 65 mA.) Each tube will be drawing this current. Now find the current allowable via the power formula: I=P/E. In our case this is 20 watts at 300 volts or 66.7 mA. Since the "opposite" tube is pulling 65 mA, subtracting this from our 66.7 ma gives 1.7 mA, which is the most (lets call it unbalanced) current we can pull. Draw a dot at 300V, 1.7 mA.
  2. Move to the next "convenient" grid line. Since, in our case, the gain is about 10, this will correspond to moving "left" by 50 volts to 250 volts. At this point the tube we have 250 volts on the upper tube, and the power formula states we could draw 20/250=80 mA. However, the "lower" tube has 350 volts on it, but is biased at -25 volts. Looking this point up on the original single ended load line indicates 57.5 mA. Subtract the 2 to obtain 22.5 mA. Draw a point at 250 volts, 22.5 mA.
  3. Repeat at 200 volts. The formula produces 100 mA. From this we subtract the "lower tubes" contribution to the current at -30 volts(grid) and 400 volts(plate) which is 45.2 mA. Subtracting this from 100 mA produces 54.8 mA. Draw a dot at 200 volts and 54.8 mA.
  4. Repeat at 150 volts (20/150=133 mA). The lower tube is biased at 450V and -35V grid (40 mA) so the maximum current is 133-40=93 mA. Draw a dot at 150V, 93 mA.
  5. Repeat at 100 volts. The formula produces 200 mA, the lower tube biases at 500V and -40V grid (32.5 mA) so the maximum current is 167.5 mA.
  6. Now connect these with a nice smooth curve.
  7. Rotate for the lower tubes power condition... -1.7mA at 300V, -22.5mA at 350V (250V on the lower tube), 54.8 mA at 400V, 93 mA at 450V and 167.5 mA at 500V and draw that condition.

You should now have a graph that looks like this...

By the way, if you have access to a spread sheet, this makes manipulating these things a LOT easier. The data for all these examples is in an Excel spread sheet. Feel free to load it if you like: Spreadsheet Data (platcurv.zip)Hint: If you replace the data in the spreadsheet with the data for the tube of your choice, most of the work is done for you, as there are embedded graphs for single ended and composite curves in the spreadsheet. There are a couple of items of interest to note in the curve above. You can definitely see the effect of transitioning from Class A to Class AB region; this is where the slope more-or-less abruptly changes in the curves above. The other thing to note is how much more linear the composite curves are than the SE case.


Push Pull Load Line

The same rules we used in the single ended case apply, with one exception: The impedance you plot is the impedance AS SEEN BY A SINGLE TUBE. Thus it is 1/4 the plate to plate load impedance. Lets use a 8800 ohm plate to plate load. In this case, each tube sees 2200 ohm load, and one "point" on the loadline is the 300 volt "0" mA quiescent value. Another convenient point is at "0" volts (namely, 300 volts drop across the 2200 ohm load). By ohms law, this is i = e/r = 300/2200 = 136 mA. As in the SE case, we need to obtain the same 5 currents and 2 voltages to give us the power output and distortion prediction. This is shown below:

Remember the formulas we used:

Po=(Ve-Va)*(Ve-Va)/(8*RL)

HD2=75*(Ia+Ie-(2*Ic))/(Ia+Ib-Id-Ie)

HD3=50*(Ia-(2*Ib)+(2*Id)-Ie)/(Ia+Ib-Id-Ie)

HD4=25*(Ia-(4*Ib)+(6*Ic)-(4*Id)+Ie)/(Ia+Ib-Id-Ie)

These work, with the RL value equal to a single tubes load (2200 ohms in our example).

The values from the graph are: Va=104V, Ve=496V, Ia=89, Ib=44, Ic=0, Id=-44, Ie=-89.

Power Out HD2(%) HD3(%) HD4(%)
8.7 watts -0- 0.38 -0-

Notice in the push pull configuration, even order distortion "cancels". Also, of more importance is the fact that the ODD order is reduced over the single ended case, even though the power output is more than double the SE case (clearly we could "parallel" 2 tubes SE and use half the impedance to get twice the power at the same distortion which would be 8.4 watts, and 0.7% third order distortion).

For those with the spreadsheet, it is left as an exercise for the student to see what happens if you don't feed the push pull stage with *exactly* the same signal level. For instance, allow the top tube to vary 0 to -20 volts to -40 volts bias, and have the bottom tube vary -35 to -20 to -5 instead of -40 to -20 to 0. Hint: you will get back some of the even order distortion products, but retain the other advantages of push pull.


The Effect of Plate Resistance on the Composite Curves

You might have noticed that the plate resistance, demonstrated on the composite curves seems to be much more constant than the single ended case. In Class A, this is definitely true. Using the tangent slope method we get:

In this case, the plate resistance is seen to vary from 430 to 550 ohms. In the SE case, RL/RP ratio varied from 2:1 to 7:1 over the operating range. In the case of the push pull arrangement, this variation is cut to 4:1 to 5:1 over the operating range. The effect of speaker load variation will be cut down as well.

I hope this little tutorial has de-mystified load line creation and manipulation. We have gone thru how to create both a single ended and push pull loadline, shown the effects of non-ideal plate resistance, and provided some formulas to allow you to predict power output and distortion from your design.

Steve

 

Part 4: The Effect of Bias Change on Composite Loadlines - from Class A through Class AB to Class B

There has been some speculation whether there is a large or small difference in the sound quality of a Class A or a Class AB amplifier. This rather short part of our series shows the effects of altering the amplifier bias. For consistency, the same "invented" tube is used.

This first curve is pure Class A, with both tubes drawing current during the entire cycle.

Notice the very constant plate resistance, and very smooth composite set. Very little distortion will be heard, whether the load is resistive (as shown), or very highly reactive (the load line looks like an ellipse instead of a straight line).


This next curve is the limit of pure class A operation. The bias has been increased (the grid biased more negative) so that one tube just reaches zero current as the other one reaches maximum current.

Notice that while the curves are still very straight, a "kink" is starting to become evident as the second tube cuts off.


The third curve is typical Class AB biasing, where the tubes are really operated in Class A for all but the highest power outputs.

Notice that there is a gradual transition towards sloppiness of the curves. There are three things to note in the Class AB operation:

  1. The plate resistance is still constant THROUGH the load line.
  2. For Reactive loads, the load will hit some really variable portions of the plate resistance. This will have audible effects if your speakers are reactive, but probably be unnoticeable on a really "resistive" speaker.
  3. There is a "common wisdom" that the plate resistance will change in Class AB when one tube turns off, since the plate resistance "changes from two parallel tubes to one, so the resistance doubles". Nonsense. The curves clearly show this does not happen. In fact, the effective plate resistance is slightly lower in those portions where one tube only is conducting.


In this next curve, the bias is increased even further. There is still significant idle current, but the transition to one tube operating occurs at much lower power levels.

In this curve, you can see fairly clearly the "crossover" region. The gain is definitely lowered for small signals. Some may think this is a form of xpansion. What it is, is a form of really obnoxious distortion. Yet for smaller signals, the amplifier is STILL running class A.


This final curve is just the threshold of Class B. One tube turns OFF just as the other one starts to conduct. This won't sound very good.


Notice there is no "magic" of the transition from one class of operation to the next, but rather a gradual transition.

Hope you've enjoyed this series.

Steve

 

 

Of Loadlines, Power Output and Distortion - Part 5 - The Pentode

Purpose

In the preceding parts of this series, we concentrated on loadlines and triode operation. In this article we will explore establishing a loadline on a pentode vacuum tube (valve), determine the power available from this loadline, and the distortion predicted from this loadline. This will allow the reader a step by step way to calculate the parameters for his design. I have again chosen to "invent" a tube for illustrative purposes. My pentode has the following parameters:


Establishing a Load Line: (very similar to a triode for S.E. applications)

  1. Get the tube characteristics for your design (shown above for my example). If not already on the characteristics, plot the maximum power dissipation curve (voltage * current = spec'd power).
  2. Select a reasonable quiescent operating point for your design. Often this is described in the manufacturers typical operation. For instance, a reasonable starting point for my example would be 300 volts and 83 mA. Whatever you pick MUST BE within the tube's ratings. Put a dot on the tube characteristics at this point.
  3. Now start drawing a potential load line. This is a straight line whose "slope" is the primary "impedance" of the transformer. How do you do this? Pick a point somewhere about DOUBLE the quiescent voltage and zero plate (anode) current. Draw a line (called "loadline" in the curve below) thru these 2 points extending to the 0 plate volt axis. Did your loadline go above the max power curve you drew? Not good. Go back, pick a different zero current point. Is the voltage at no current more than twice the "rated" voltage? Not good. Go back and pick another point. Within spec? Good. Now determine the impedance. Z=(maxvoltage-quiescent voltage)/quiescent current. Alternately, you can pick an impedance to suit available plate transformers. In this case, if you go outside the safe boundaries, you must choose a new quiescent point. This is where picking a loadline for the triode and the pentode differ. Maximum power output and more-or-less minimum distortion for the pentode occurs where the load line goes through the "knee" of the curve. This is illustrated above for a load of 2.4k. Notice I have "exceeded" the tubes rating over a small portion of the curve. You can do this, as the power dissipation over the full conduction cycle is still within the tube rating. Make sure the idle condition is within the ratings, however.
  4. Having now chosen a tentative loadline, we need the following information from it......
    1. The required grid bias (you'll need to supply the grid with this value of DC voltage).
    2. The quiescent voltage (called Vq)
    3. Voltage where the grid bias voltage is zero (called Va).
    4. Voltage where the grid bias is double the quiescent value (called Ve). Note... if the plate current reaches zero before you get to this bias, you've got a bad operating point, go back to step 3 and try again.
    5. Current at 0 volt grid bias (called Ia). If this current is above the maximum rated tube current, go back to step 3 and try again. Hint: If its off the scale, chances are the current is too high.
    6. Current at half the quiescent bias voltage. (Called Ib.)
    7. Quiescent Current.(Called Ic.)
    8. Current at 1.5 times the quiescent bias voltage (Called Id).
    9. Current at twice quiescent bias voltage. (Called Ie).

    For our example, the data is as follows:

    Va= 70V, Vq=300V, Ve=433V, Ia=180mA, Ib=130mA, Ic=83mA, Id=51.5mA, Ie=28mA. In addition, 40V p-p are required from the driver stage, and a 2.4k plate transformer is required.


    Rectification Effects

    In any device with even order distortion (not just second harmonic as sometimes stated), the average current of a Class A stage will change depending on the signal level. If the distortion is relatively low, the effect is relatively unimportant, if the distortion is high, this effect becomes increasingly important. (It is also more important for self bias than fixed bias, as the bias voltage is a function of the average current).

    This effect modifies all tube operating characteristics. The degree can be seen by comparing the quiescent current (83 mA in our example) with the average current as taken from the "extremes": that is, (Ia+Ie)/2. In our example that value calculates to 104mA (note that since this is reasonably different, we can expect relatively high distortion). The effect requires an iterative plot of the load line on the tube characteristics as described in part 1 of this series. Note the quiescent dissipation is still 300*0.083, or in this case 24.9 watts. The non-linearity shown here is getting dangerously close to what can be ignored, but, as we will see, the distortion in this example is too high anyway.


    Power Output

    Po = (Ve-Va)*(Ve-Va)/(8*loadimpedance).

    In our example this is (433-70)*(433-70)/(8*2500)= 6.9 watts. This number is an approximation in that it assumes low distortion.

    Second Harmonic Distortion

    HD2(%) = 75*(Ia + Ie - 2*Ic)/(Ia + Ib - Id - Ie)

    In our example this is 14% (!)

    Third Harmonic Distortion

    HD3(%) = 50*(Ia - (2*Ib) + (2*Id) - Ie)/(Ia + Ib - Id - Ie)

    In our example this is -1.1%

    Notice the minus sign. This indicates that the harmonic content subtracts from the fundamental (flattening it) when the fundamental is at its crest. This *usually* happens on third harmonic distortion in tubes.

    Fourth Harmonic Distortion

    HD4(%) = 25*(Ia - (4*Ib) + (6*Ic) - (4*Id) + Ie)/(Ia + Ib - Id - Ie)

    In our example this is -2.2%

    It is also possible to get an indication of how distortion varies with power output. In the example we just used, we picked 0, -10, -20, -30 and -40 volt bias points. We could also use -10, -15, -20, -25 and -30 points to see what happens at lower levels.

    Applying the same formula to these points gives us 187 and 376 volts, and 130, 110.5, 83, 69.5 and 51.5 mA. This, in turn, calculates to 1.9 watts and 9.7% second harmonic, -1.5% third harmonic, and -8.5% fourth harmonic. This means the 2 watt level with this tube will also be pretty distorted!


    Push Pull Operation (composite loadlines for pentodes)

    We will discuss how to create a composite load line for a pentode. This is "necessary" to establish the load line for a push pull amplifier, regardless of whether it is going to be Class A or Class AB. We will continue to use the same "invented" pentode. As we will discover, establishing a realistic loadline for a pentode is slightly different than for a triode, due to the shape of the characteristic curves.

    Building a Composite Load Line

    This is again an iterative process, perhaps moreso than the single ended case as it's a bit more involved. Here's a step by step procedure to create a set of "composite curves" and load line.

    1. Establish your intended operating point from the single ended curves. First let's consider an obvious Class A case, by choosing the same bias point we used for the single ended case: 300 volts and 83 mA quiescent current (-20 volt bias). Note: For Class AB, you may choose a lower idling current point, which will allow a lower impedance loadline (and higher power output) without exceeding maximum allowable power dissipation. For instance, choosing a -25 or -30 volt quiescent point would allow you more "room" to increase tube current. To stay within Class A operation, at least some current must flow in each tube at all bias points, otherwise, you will be operating in Class AB. In this article we will consider -20 volt bias point (requiring a swing on each of the grids of 0 to -40 volts, -25 volts (which is still technically class A, since there is some current still flowing at -50 volts bias) and -30 volts, which is Class AB, since no current flows at -60 volts bias.This is actually an important distinction with pentodes. Since the "cutoff" characteristics tend to be sloppier than a triode, most so-called Class AB amplifiers are really Class A, as some small current is flowing even at the most negative point. As we will see, truly Class AB has some pretty bad crossover (notch) distortion.

    2. Since the plate voltage increases on one plate and decreases on the other plate, we must have a way of representing this. The usual method is to take another set of the same tube curves, turn them around (so the maximum current is "down" and highest plate voltage is to the "left", and position them so that the quiescent voltages line up vertically, and the zero plate current lines touch each other as shown below. I've shown the steps to do this on the curves below: that is, copy the curves, turn one around, then merge the curves as:

    3. Now, for the chosen bias point, create a new line (labeled "composite" curve) as follows (we are going to look at ONLY the -20 volt bias lines, one on the "upper" part of the graph, one on the "lower" part of the graph:

  5. At the quiescent point (300V), subtract the currents (83-83=0 mA). Draw the "dot" at 300V, 0 mA on the composite curves.
  6. Step to the next convenient voltage, say 310/290 volts, and subtract the currents. Plot this point.
  7. Step to the next convenient voltage, say 320/280 volts, and subtract the currents. Plot this point. Continue until you have the -20 volt bias point established.

4. You have now established one "line" of the composite curves. This is the line of varying the plate voltage symmetrically about the quiescent point while maintaining a constant grid voltage. (classic plate resistance line).

5.  Now consider another set of lines for your graph. This will be the -25 volt line for the "upper" set and the -15 volt line from the lower set. (We have put a 5 volt signal into the push pull stage). Again, for each plate voltage, subtract the two currents (for instance, 310V -25V bias upper and 290 volt -15V lower) and plot the dot. Continue until you have the -25/-15V line filled in. Note that if your quiescent grid voltage was, for instance, -30 volts, the second line you would add would use the -35/-25V lines instead of -25/-15V line.

6.  Next do the same for the -30/-10, the -35/-5 and -40/0 volt lines. Then do the same for the -15/-25, the -10/-30, the -5/-35 and the 0/-40 volt lines. This completes the composite curves. The last 4 curves should be mirror images of the previous 4 you filled in. Noticing this saves you some time.

7. The last thing to do is fill in the maximum power dissipation curve. There are now going to be 2 of them, one for the one tube, one for the other. These will be symmetrical about the quiescent point. It is usually necessary to graph only half of the curve. The following is for the "upper" tube. Here's how to do it:

  1. Get the current at the quiescent point. (In our case that's -20 volts, 300V, 83 mA.) Each tube will be drawing this current. Now find the current allowable via the power formula: I=P/E. In our case this is 25 watts at 300 volts or 83.3 mA. Since the "opposite" tube is pulling 83 mA, subtracting this from our 83.3 ma gives 0.3 mA, which is the most (lets call it unbalanced) current we can pull. Draw a dot at 300V, 0.3 mA.
  2. Move to the next "convenient" grid line. This will correspond to moving "left" by about 60 volts to 240 volts. At this point the tube we have 240 volts on the upper tube, and the power formula states we could draw 25/240=104 mA. However, the "lower" tube has 360 volts on it, but is biased at -25 volts. Looking this point up on the original single ended load line indicates 65 mA. Subtract the two to obtain 39 mA. Draw a point at 240 volts, 39 mA.
  3. Repeat for a series of grid conditions just like we did for the triode case.
  4. Now connect these with a nice smooth curve.
  5. Rotate for the lower tubes power condition: it's a mirror image remember.

You should now have a graph that looks like this...

Push Pull Load Line

The same rules we used in the single ended case apply, with one exception: The impedance you plot is the impedance AS SEEN BY A SINGLE TUBE. Thus it is 1/4 the plate to plate load impedance. Lets use a 6000 ohm plate to plate load. In this case, each tube sees 1500 ohm load, and one "point" on the loadline is the 300 volt "0" mA quiescent value. Another convenient point is at "0" volts (namely, 300 volts drop across the 1500 ohm load). By ohms law, this is i = e/r = 300/1500 = 200 mA. As in the SE case, we need to obtain the same 5 currents and 2 voltages to give us the power output and distortion prediction.

Remember the formulas we used:

Po=(Ve-Va)*(Ve-Va)/(8*RL)

HD2=75*(Ia+Ie-(2*Ic))/(Ia+Ib-Id-Ie)

HD3=50*(Ia-(2*Ib)+(2*Id)-Ie)/(Ia+Ib-Id-Ie)

HD4=25*(Ia-(4*Ib)+(6*Ic)-(4*Id)+Ie)/(Ia+Ib-Id-Ie)

These work, with the RL value equal to a single tubes load (1500 ohms in our example).

The values from the graph are: Va=71V, Ve=529V, Ia=151, Ib=77, Ic=0, Id=-77, Ie=-151.

Power Out HD2(%) HD3(%) HD4(%)
17.5 watts -0- 0.7 -0-

Notice in the push pull configuration, even order distortion "cancels". Also, of more importance is the fact that the ODD order is reduced over the single ended case, even though the power output is more than double the SE case. Even more important is the significance of the picked load. Again, we picked a loadline that goes through the "knee" of the composite curves. We will show below, that as the bias conditions are altered, the "best fit" impedance changes also, which is quite a change from the triode case.

This is one case where the operation of these two device types is quite different.


The Effect of changing the bias on Pentode Composite Curves

Let's choose a -25 volt bias on the tubes, which will drop the idle current to 60 mA. As I mentioned above, technically, this is still Class A operation. The curves look like this:

One important item to note is that now, an appropriate load line is 5.2k instead of 6k. (Substantially lower load impedance produces lower power and higher distortion, as does substantially higher load impedance.)

For this condition, The seven values picked from the curve are 74V, 526V, 174, 83, 0, -83 and -174 mA.

Power Out HD2(%) HD3(%) HD4(%)
19.6 watts -0- 1.6 -0-

Notice the available power output has increased, but so has the distortion.

This is even more obvious if we move the class of operation into true Class AB:

Now, the appropriate loadline has decreased to 5k, but the power output has reached 20 watts.

We can pick 0, -15, -30, -45 and -60 volt points (which gives us full 20 watts output) or choose intermediate points. This is summarized in the following table:

Power Out HD2(%) HD3(%) HD4(%)
20 watts -0- 2.5 -0-
8.9 watts -0- 5 -0-
1.5 watts -0- 3.4 -0-

Now you can see the effect of the "compressed" curves around zero bias. The distortion is actually greater at low levels than at maximum output. This is the effect of crossover or notch distortion.

 -Steve

 

 

 

What Harmonic Distortion "looks like".

This short tutorial is intended to explain what different kinds of distortion look like (as seen on an oscilloscope). Although by no means is this an "exhaustive" display of every type of distortion vs amplitude and phase of harmonics, this is representative of the type of waveforms you are likely to encounter. By studying the displays shown in this article, you can get a reasonable idea of what the distortion in your amplifier is, by comparison.

Just to calibrate things, so to speak, the first picture is a pure sinusoidal wave:

Notice the nice smooth symmetry of this waveform.

Next is a waveform with exactly 5% second harmonic added:

If you look carefully, you will see that the top of the waveform is narrower, and the bottom of the waveform is wider than the true sine wave. The degree of asymmetry is correlated to the distortion. Note that the waveform is still smooth, as there is ONLY second harmonic distortion present in this waveform.

Odd order distortion, on the other hand, is symmetrical around the "middle" as shown in this waveform that has exactly 5% third harmonic distortion:

What you should notice here is both the top and bottom of the waveform are fatter.

When higher order distortion components are present, their effect becomes slightly more visible on the scope. Here's a waveform consisting of 5% second, 2% fourth and 1% sixth harmonics. Notice this represents ONLY even order distortion.

Comparing this with the second harmonic ONLY case, you can notice the top is definitely sharper, and the bottom definitely flatter. The "sharpness" is a give-away that there's higher order distortion components.

The same kind of thing happens with odd order distortion. Here is a waveform with 5% third and 2% 5th order distortion products:

Again, the symmetry is a giveaway that the distortion is ODD order, and the flatness of the top and bottom is a giveaway that there's higher order components.


Many real world amplifiers have both ODD and EVEN order distortion products. Here is a waveform with 5% second, 2% third, 1% fourth, 0.5% fifth and 0.2% sixth harmonics. This might be representative of an amplifier reaching its maximum output:

Interestingly, this waveform doesn't LOOK too bad. Lets reverse the amplitudes, just to compare things. In this image, the harmonic distortion is 0.2% second, 0.5% third, 1% fourth, 2% fifth and 5% sixth harmonc:

Notice that the THD reading of these last two figures would be the same, yet the bottom one LOOKS (and would sound) much worse. This is one of the reasons why a decreasing spectrum content amplifier sounds better than one in which the higher harmonics predominate, even though the latter might have "lower THD".

I hope this gives you some idea what different distortion components LOOK like, so you can judge what your amplifier is producing. This also allows you to correlate what you hear with what you see.

Technical Note: The waveforms here were created from the form:

   y = cos(x) + %2nd*cos(2x) - %3rd*cos(3x) + %4th*cos(4x) - %5th*cos(5x) + %6th*cos(6x).

Steve

 

 

Operating Characteristics of different connections of the Pentode

This report provides typical tube characteristics using different Pentode connections. This will be done by showing the resulting plate characteristic curves of these different connections. Standard Pentode connection (g1 as stepped parameter, g2 at a constant voltage, and plate voltage and current plotted for the stepped input), triode connection (g2, g3 connected to the plate) and suppressor input triode mode (g1=0V, g2 at a constant voltage, g3 as the stepped (input) parameter) are presented. A 6888 device is used as the example.

Pentode mode

This presents a relatively high impedance (constant current) output impedance. The relatively sharp knee at "0V grid bias" provides relatively high available power output for a given plate supply voltage. Note, however, that the knee of this "true pentode" is not as sharp as a beam power tube is, though. There is also a 100 ohm screen grid "stopper" resistor in series with the 90V source. This also contributes to softening of the knee just a bit.

G3 driven

In a true pentode (as opposed to a beam power tube), depending on the construction, it is sometimes practical to use the suppressor grid as the signal input. This provides triode like characteristics as there are no intervening control structures between it and the plate. The MU of the resulting triode like device is high. Sometimes this connection, because of the relatively high MU allows a reasonable power with good sensitivity. In the curve below, the tube is configured with g1 grounded (0 volts) and the screen at a constant 90VDC. The 6888 is particularly well suited to this application, being a "dual control" device. Notice the maximum available current is similar to that of the pentode connection, but the zero bias point is slightly lower. G1 was grounded via a 10k resistor, and this provides slightly less than 1 volt negative bias to G1, due to the "contact bias". If this point were hard grounded, the "zero bias" line (top red) would be identical to that of the pentode mode connection., and the other curves would shift up slightly. Note the MU is 35-40. One additional item of interest is the fact that some current is available at "zero plate volts", due to the G2 acceleration. Like the pentode mode, there is a 100 ohm screen stopper series resistor.

Triode mode

In this connection, g2 and g3 are connected to the plate, and the device is operated as a triode. For this connection, the MU is generally fairly low, and the corresponding plate resistance is relatively low as well. This particular "pentode" (6888) is amazingly linear in this mode of operation. Note the low MU (about 8), and rp below 1k ohm. Also note that the 90 volt zero bias point is higher than the 90 volt plate, 90 volt screen point of the pentode connection. This is due to the use of the screen stopper in the pentode and g3 driven case and the fact that the screen in those two modes does not contribute to the overall "plate" current as it does in triode mode.

There are other connections possible with a pentode, three of which are reasonably popular. One of these is "ultra linear", generally done by way of a tap on the transformer. In this connection, the voltage on g2 is some portion of the plate voltage. The characteristics are intermediate between triode and pentode connection.

The second sometimes used connection is "enhanced triode" connection. In this connection, the input is on G2, generally between 0 and near the plate voltage. The power gain of this connection is low, since G2 consumes current, as it is forward biased.

The third connection is space charge grid connection. Here g1 has some fixed positive bias, the input is a negatively biased g2, acting as the control grid, and g3 connected to plate. This allows a very high relative current as electrons are pulled from the cathode by the nearest available structure. Unfortunately, this is also fairly wasteful of cathode current, and not very linear.

Expansion on these items may become another article.

Steve

 

 

 

Grounded Cathode Amp.

Grounde and safety earth.

http://www.tubecad.com/may2000/index.html

http://electronluv.com/index.php

 

Meget mere info kommer...

AUDIO TRANSFORMERS